If $K \triangleleft G$ and $P$ a Sylow p subgroup of a finite group $G$, then $K \cap N_G(PK)=\left\{e\right\}?$ I read something, see 
Marshall Hall Jr., On the number of Sylow subgroups in a finite group, Journal of Algebra
Volume 7, Issue 3, December 1967, Pages 363–371 DOI, Theorem 2.1  
"The inverse image of $N_H(P^*)$ is $N_G(PK)$"
that suggested the following: If $K \triangleleft G$ and $P$ a Sylow p subgroup of a finite group $G$, then $K \cap N_G(PK)=\left \{ e \right\}$, where $N_G$ stands for the normalizer in $G$. Is this true? If so, does $P$ have to be a Sylow p subgroup of $G$?
 A: Perhaps these observations help. All groups mentioned are finite.
Proposition 1 Let $H \leq G$ and $N \unlhd G$, with $N \subseteq H$. Then 
$$N_G(H)N/N=N_{G/N}(H/N)$$
Proof This relies heavily on the so-called correspondence theorem, moving back and forth between subgroups of a group and its quotients w.r.t. a fixed normal subgroup (see for example I.M. Isaacs, Finite Group Theory, X.21 Theorem).Since $H \unlhd N_G(H)$, we have $H/N \unlhd N_G(H)N/N$ and hence $N_G(H)N/N \subseteq N_{G/N}(H/N)$. For the converse containment put $N_{G/N}(H/N)=U/N$, with $U$ a (unique) subgroup of $G$ containing $N$. Now $H/N \unlhd N_{G/N}(H/N)$, so $H \unlhd U$, whence $U \subseteq N_G(H)$, so $U/N \subseteq N_G(H)N/N$ and we are done.
Corollary 1 Let $H \leq G$ and $N \unlhd G$, then
$$N_G(HN)N/N=N_{G/N}(HN/N).$$
Proof In Proposition 1 instead of $H$, put this subgroup to be $HN$.
Observe that always $N_G(H)N \subseteq N_G(HN)$. Now, one probably tends to think that $N_G(HN)=N_G(H)N$, but in general this is not true. 
 However, if $H$ is a Sylow $p$-subgroup, then we have equality.
Proposition 2 Let $P \in Syl_p(G)$, $N\unlhd G$, then
$$N_G(PN)=N_G(P)N.$$
Together with Corollary 1, this yields the following.
Corollary 2 Let $P \in Syl_p(G)$, $N\unlhd G$, then
$$N_G(P)N/N=N_{G/N}(PN/N).$$
Proof of Proposition 2 Note that $N_G(P) \subseteq N_G(PN)$ since $N$ is normal. Observe further that $P$ is Sylow in $N_G(PN)$, in fact we have $P \subseteq PN \unlhd N_G(PN)$, and we can apply the Frattini Argument: $$N_G(PN)=N_{N_G(PN)}(P)PN=(N_G(PN) \cap N_G(P))PN=N_G(P)PN=N_G(P)N,$$ concluding the proof.
